y^{''}=y,y(0)=0,y^'(1)=1

Lösung

y^{^{''}} = y, y (0) = 0, y^{^{'}} (1) = 1

y = - \frac{e}{- e ^{2} - 1} e^{t} + \frac{e}{- e ^{2} - 1} e^{- t}

Schritte zur Lösung

y^{''} (t) = y

Solve linear ODE:

y = - \frac{e}{- e ^{2} - 1} e^{t} + \frac{e}{- e ^{2} - 1} e^{- t}

y = - \frac{e}{- e ^{2} - 1} e^{t} + \frac{e}{- e ^{2} - 1} e^{- t}

\frac{d ^{2} y}{d t ^{2}} - 11 \frac{d y}{d t} + 10 y = 0

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = - 2, y^{^{'}} (0) = 21

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 4, y^{^{'}} (0) = 15

9 y^{^{''}} + 8 y^{^{'}} - 7 y = 0

y^{^{''''}} - 3 y^{^{'''}} - 6 y^{^{''}} + 28 y^{^{'}} - 24 y = 0